Problem: The lifespans of turtles in a particular zoo are normally distributed. The average turtle lives $110$ years; the standard deviation is $22$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a turtle living less than $132$ years.
Solution: $110$ $88$ $132$ $66$ $154$ $44$ $176$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $110$ years. We know the standard deviation is $22$ years, so one standard deviation below the mean is $88$ years and one standard deviation above the mean is $132$ years. Two standard deviations below the mean is $66$ years and two standard deviations above the mean is $154$ years. Three standard deviations below the mean is $44$ years and three standard deviations above the mean is $176$ years. We are interested in the probability of a turtle living less than $132$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the turtles will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the turtles will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $88$ years and the other half $({16\%})$ will live longer than $132$ years. The probability of a particular turtle living less than $132$ years is ${68\%} + {16\%}$, or $84\%$.